Question

find a point on x axis which is at a distance of 5 units from (4,6)​

Answer 1

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FORMULA TO BE IMPLEMENTED

1.

Any point on x axis is of the form ( a , 0 )

Where a is a real number

2.

Distance between given two points

$(x_1,y_1) \: and \: \: (x_2,y_2) \: is$

$= \sqrt{ {(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2} } \: \: \: \: unit$

TO DETERMINE

A point on x axis which is at a distance of 5 units from (4,6)

CALCULATION

Let ( a, 0) be the point

Then by the given condition

$\sqrt{{(a - 4)}^{2} + {(6 - 0)}^{2} } = {5}^{2}$

$\implies \: {(a - 4)}^{2} = 25 - 36$

$\implies \: {(a - 4)}^{2} = - 11$

There exists no such real number such that the square of the number is negative

So there is no such point on x axis which is at a distance of 5 units from (4,6)